[Toán 7] Tìm x

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nhuquynhdat

$\dfrac{x+5}{200}+\dfrac{x+4}{201}=\dfrac{x+3}{202 }+\dfrac{x+2}{203}$

$\leftrightarrow \dfrac{x+5}{200}+1+\dfrac{x+4}{201}+1=\dfrac{x+3}{202 }+1+\dfrac{x+2}{203}+1$

$\leftrightarrow \dfrac{x+205}{200}+\dfrac{x+205}{201}-\dfrac{x+205}{202 }-\dfrac{x+205}{203}$

$\leftrightarrow (x+205)(\dfrac{1}{200}+\dfrac{1}{201}-\dfrac{1}{202 }-\dfrac{1}{203})$

Do $\dfrac{1}{200}+\dfrac{1}{201}-\dfrac{1}{202 }-\dfrac{1}{203} \ne 0$

$\Longrightarrow x+205=0 \leftrightarrow =-205$
 
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